Given, PQ II RS II UT
(a) Given, c = 57° and \(a=\frac{c}{3}\)
\(\because PQ||UT\)
\(\therefore \angle UTP=\angle QPT\) [alternate interior angles]
\(\Rightarrow \angle c=\angle a+\angle b[\because QPT=a+b]\)
\(\Rightarrow 57^0=\frac{57^0}{3}+\angle b\)
\(\Rightarrow 57^0-19^0=\angle b\)
\(\Rightarrow \angle b=38^0\)
\(\therefore \angle b+\angle d=180^0\)
\(\Rightarrow \angle d=180^0-38^0=142^0\)
(b) Given, c = 75° and \(a=\frac{2}{3}c\)
\(\Rightarrow c=75^0\) and \(a=\frac{2}{5}\times75^0=30^0\)
\(\therefore \angle c=\angle a+\angle b\) [alternate interior angles]
\(\Rightarrow 75^0=30^0+\angle b\)
\(\Rightarrow 75^0-30^0=\angle b\)
\(\Rightarrow \angle b=45^0\)

Given,\(\angle 1=(2a+b)^0\) and  \(\angle 6=(3a-b)^0\)
so, \(\angle 1=\angle 7\) [alternate exterior angles]
So, we have \(\angle 7=(2a+b)^0\)
\(\therefore \angle 6+\angle 7=180^0\)              [linear pair]
\(\Rightarrow\) (3a-b)⁰ + (2a + b)⁰ = 180°
\(\Rightarrow\) 3a- b + 2a + b = 180°
\(\Rightarrow\) 5a = 180°
on dividing both sides by 5, we get
\(\Rightarrow a=\frac{180^0}{5}=36^0\)
\(\because \angle 1+\angle 2=180^0\)                    [linear pair]
\(\Rightarrow 2a+b+\angle 2=180^0\)
\(\Rightarrow 2\times36^0+b+\angle 2=180^0\)
\( \Rightarrow b+\angle 2=180^0-72^0\)
\(\Rightarrow b+\angle 2=180^0\)
\(\therefore \angle 2=(108-b)\)

\(\angle BAD=\angle BAE+\angle EAD\)
= 40° + 30°
= 70°
and \(\angle CDA=70^0\)
\(\therefore \angle BAD=\angle CDA\)
But they form a pair of alternate angles.
\(\Rightarrow\) AB II CD .........(i)
Also, \(\angle BAE+\angle AEF=40^0+140^0=180^0\)
But they form a pair of interior opposite angles.
\(\Rightarrow\) AB II EF..........(ii)
From (i) and (ii),we get
AB II CD II EF
\(\Rightarrow\) CD || EF

(a) Since EF II GH and AC is transversal.
\(\Rightarrow \angle CAF+\angle ACH=180^0\)
(interior angles on same side of transversal)
\(\Rightarrow \angle CAF=180^0-105^0\)
=75⁰
(b) Since EAF is a straight line.
\(\therefore \angle EAB+\angle BAC+\angle CAF=180^0\)
\(\Rightarrow 60^0+\angle BAC+75^0=180^0\)
\(\Rightarrow \angle BAC=180^0-135^0=45^0\)

Angle between d and e +
Angle between e and c = 180°
[by linear pair]
\(\therefore\)  Angle between d and e=180°-75°=105°
[\(\because\) Angle between c and e = 75°]

Angle between band c = 30°
[vertically opposite angles]

Since, lines AB, CD and EF intersect each other at point O.
Where, \(\angle\)AOE = 30° and \(\angle\)DOB = 40°
\(\because\) \(\angle\)AOE and \(\angle\)FOB are vertically opposite angles.
\(\therefore\) \(\angle\)FOB= 30°
Also, \(\angle\)DOB + (\(\angle\)BOF + \(\angle\)FOC) form a linear pair.
So, \(\angle\)DOB + \(\angle\)BOF + \(\angle\)FOC =180°
\(\Rightarrow\) 40° + 30° + \(\angle\)FOC =180°
\(\Rightarrow\) 70° + \(\angle\)FOC =180°
\(\Rightarrow\) \(\angle\)FOC = 180° - 70°
\(\Rightarrow\) \(\angle\)FOC =110° or \(\angle\)COF =110°

Since, AF II ED and EF is a transversal line.
So, \(\angle \)AFE = \(\angle \)FED = 40° [alternate interior angles]
Also, AB II CD
70° + (\(\angle \)AFE + \(\angle \)EFD) form a linear pair angles

Since, \(\angle\)AOB = 90° [right angle at O]
So, \(\angle\)AOB = \(\angle\)COB + \(\angle\)COA
\(\because\) \(\angle\)COB = 49° \(\Rightarrow\) \(\angle\)COA + 49° = 90° [given]
\(\Rightarrow\) \(\angle\)COA = 90° - 49° \(\Rightarrow\) \(\angle\)COA = 41°
Since, DC is a straight line, where \(\angle\)COA and \(\angle\)AOD form a linear pair .
So, \(\angle\)AOD + \(\angle\)COA =180°
\(\because\) \(\angle\)COA =41°
\(\therefore\)  \(\angle\)AOD =180° - 41° \(\Rightarrow\)  \(\angle\)AOD =139°

Given, p || q and I is a transversal.
We know that, the sum of pair of interior angles on the same sides of the transversal is supplementary.
\(\therefore\) \(\angle \)a + 60° = 180° \(\Rightarrow\) \(\angle \)a = \(\angle \)180° - 60° = 120°
and I || m and q is a transversal.
\(\therefore\) \(\angle \)a = \(\angle \)1          [pair of corresponding angles]

\(\angle \)1 = 120°
and \(\angle \)d= \(\angle \)1= 120°    [vertically opposite angles]
Now, \(\angle \)1 + \(\angle \)c= 180°
[\(\because\) sum of the angles on the same side of a transversal is 180°]
\(\Rightarrow\) 120° + \(\angle \)c = 180° \(\Rightarrow\) \(\angle \)c = 180° -120° = 60°
\(\Rightarrow\) \(\angle \)b = \(\angle \)c = 60° [vertically opposite angles]
Hence, \(\angle \)a = 120°, \(\angle \)b = 60°, \(\angle \)c = 60° and \(\angle \)d=120°